hi Folks,

I've made a simple voltage divider to bring 350VDC down to a max of 3.3 for measuring. I've done this with 3 470k resistors (0.6W) and a 10K to ground. Across the 10K I also have a 3.3 zener in the (vain?) hopes that should something happen to one of the 470s, it would clamp the voltage.


I've built this on vero board and removed the tracks (heat, soften glue, rip off) between the resistors (two tracks spacing).

Is there anything else I should be doing to remain / keep my netduino save? Resistors are rated at 250V each.

Should I have something the other way round? i.e. if the 10k resistor goes open (can it happen?) I still see the zener taking the brunt if that happens?

Would the power dissapated by the zener be (Vin - 3.3) * I or V*I? it's only only a 1/4W. The way I see it is if one of the 470s shorts, I will have a Vin of over 5V @7uA which is still very low.


questions questions questions


TIA


Cheers,
Crispin

 

Hard to comment without seeing the layout, but 350 volts is not very high.

The breakdown voltage of air is 33KV/cm but this is between perfectly rounded terminals.
It can be very much less if there are any points, corners, elbows, solder whiskers etc.

The important thing to realise is that if you place resistors side by side and connect alternate adjacent ends to form a chain, the non connected ends have twice the voltage across any one resistor between them.
For this reason such chains are often made with the resistors diagnal or at right angles, for example in 1 -2 KV chains in scopes.

go well

 

thanks - bit confused by your caution though.
I have them end on end running 90 degrees to the vero strips. (Across the board as apposed to along it?) I've removed the tracks between the resistors.

is there something else I should be adding / doing differently to help protect my netduino or is the chance of the divider going south so remote that the zener is enough.
Considering it _only_ costs about £20, if it does fry it's not like I am loosing millions...

 

I can't remember a single case where a resistor had a short circuit, if something goes wrong they usually increase in resistance value or open. The only thing that can happen is that you, by doing in circuit measurements somehow short one of the resistors.

Your zener voltage is also very near to the voltage value you are measuring. Are you sure it doesn't distort your measurement?

If you really want to use a zener diode, you could use the next bigger one to make sure it doesn't have any influence.

If you wanted to be absolutely safe, you could isolate the high voltage with a circuit using a linear optocoupler like the HCNR201, but then you'd need a second isolated power supply for the primary side opamps...

 

If any of the 470's go short not much happens to the input, even if 2 short it only goes to 7.3 volts. However, it will deliver between .24 and .74 amps when they start to short. That current goes into your zener so it needs to dissipate .74*3.3=2.5 watts, which is a lot of power.

You may be better served by using a big power diode to V+. That only needs .74*.7 = .5 watts worst case. Read your spec sheet, this might have to be a Shockley diode to keep the voltage below the max without conduction thru the internal diode (which WILL fail).

The 10K opening is a likely and very bad failure mode.

I would prefer using higher resistors here to limit the currents, but you probably picked 10K to match the analog input. Going higher would need a buffer (op amp?).

I do not recommend using an opto couple for analog measurements. They are basically digital devices.

 

I do not recommend using an opto couple for analog measurements. They are basically digital devices.

Hi Ernie, in case you were referring to my post, I was talking about a linear analog optocoupler. I used it several times for exactly this purpose, isolated measuring of 250 to 470VDC. Works very well.

 

If any of the 470's go short not much happens to the input, even if 2 short it only goes to 7.3 volts. However, it will deliver between .24 and .74 amps when they start to short. That current goes into your zener so it needs to dissipate .74*3.3=2.5 watts, which is a lot of power.

You may be better served by using a big power diode to V+. That only needs .74*.7 = .5 watts worst case. Read your spec sheet, this might have to be a Shockley diode to keep the voltage below the max without conduction thru the internal diode (which WILL fail).

The 10K opening is a likely and very bad failure mode.

I would prefer using higher resistors here to limit the currents, but you probably picked 10K to match the analog input. Going higher would need a buffer (op amp?).

I do not recommend using an opto couple for analog measurements. They are basically digital devices.

With three 470KΩ resistors in series and 350V input, the chain current remains below 1mA unless all three resistors have failed to low resistance. Where do you get 0.74 Amps from???

 

Do you have protection for reverse polarity (-350V)?

 

thanks for the replies guys -

as for the zener being 3.3, I did not think that it would be a problem? The max I would be reading is about 2.8V @ 350, in reality my max working is around 290V. Do they reliably "operate(?)" at the rated voltage? The reason I chose 3.3 is because the Netduino cannot accept more than 3.3. Datasheet has it at vMax. Am I cutting it too close?

Seeing as I have built this one, and fully accept it is a draft version, I will take all into consideration (and be back for more questions ) for the next version.

ErnieM - how could you mitigate the 10k failing? Putting more higher ones in parallel so one failure has a smaller affect?
Having higher resistance on the high side would certainly lower the currents, is there any benefit in using say 12M / 100k (each side x 10). Would it become noisy? I think my option is 2uA - is that considered high in a situation like this?

questions questions question



Thanks again folks.

 

Do you have protection for reverse polarity (-350V)?

Urm, no.
Would the fact that the zener now becomes a normal diode not save the day?

 

Urm, no.
Would the fact that the zener now becomes a normal diode not save the day?

Yes, so long as it does, and so long as the input you are protecting can tolerate a small -ve bias of -0.7 Volts or so.

 

Will check - thanks.

 

One last thing: Will you have high voltage transients on your 350V? If so these may appear in an attenuated form on your 10k resistor. In this case you may consider putting a small ceramic capacitor in parallel with the 10k. This will of course put a small delay (us to ms) in your measurement...

 

Thanks - will do.

Unlikely that it would happen (or would it?) as the source is a PV array.

 

Thanks - will do.

Unlikely that it would happen (or would it?) as the source is a PV array.

ok, I didn't know that. I'm sure it can happen caused by nearby lightning (wouldn't destroy the PV-array nor the inverter/charger but could cause transients).
If solar inverters have filters at the input to avoid causing voltage transients on the high input voltage I cannot say, maybe someone else has an answer to that?

not directly related as it is to protect the PV power electronics: http://www1.eere.energy.gov/solar/pdfs/10_ropp.pdf

If you want to be on the absolute safe side you may have to do further research..